2007-01-15

Bang Bang Bang

.

First of all, just let me say that Monday 23 might be the greatest album ever recorded. And thus Sunfall Festival requires your full attention, so listen up:

MONDAY 23 MAY BE THE GREATEST ALBUM OF ALL TIME

Since they recorded it after they won the $250,000 contract from GarageBand, many people (myself included, at the time of purchase) assumed that they had made the album with that money. Not so. In fact (as I learned when researching my Wikipedia article), they never even ended up taking the contract. But they did use the connections to get a highend producer to work with them on Bang Bang Bang, which I purchased as soon as was humanly possible after its release.



And as the nature of humanity, it could not live up to its predecessor. It was a good album, sure, but it was no Monday 23. And whereas the previous album made excellent use of a long empty space, Bang Bang Bang's big lull is pointless and annoying.

This weekend I may be wifefree, but by no means will I thus be chickfree. So I rummaged through our still unpacked boxes of cds and found Bang Bang Bang.

Lead singer Amy is high on my list of Great Chick Voices and so she has been accompanying my travails the last three days with Bang Bang Bang on endless repeat.

And no, it's not as good as Monday 23 (even the Beatles only have one album that can really compete), but it is an incredible album and you know what? I love love love Bang Bang Bang.

I should have written this post when I first got Bang Bang Bang and 23b in the mail, when I could go on and on and on about Bang Bang Bang's qualities and shortcomings (because it was all I thought about), but maybe this is better. I will just celebrate its excellence and let you buy your own copies and share in the goodness.

Just one question: Any idea when they're playing the East Bay?

11 comments:

  1. Because of your review here, I had to look the Monday 23 up and find some sound files. I like it. It reminds me of Feist in a way. Do you listen to Feist? If not I think she would make a nice addition to your chick music collection.

    Oh, and we got to speak to the Big O on the phone last night. What a CUTIE!!!! A total crack up.

    ReplyDelete
  2. .

    I'm so glad he talked to you! Did he know who you were?

    I've had websites recommend Feist to me as well, but now that I human being I know has as well, I'll have to prioritize it.

    Sunfall Festival Fans should also look into The Sundays--Static and Silence may not be as good as Monday 23, but it might be the most beautiful album ever recorded.

    I'm totally an expert on superlatives.

    ReplyDelete
  3. My music guru brother was the one who first told me about her.

    MSB was actually the one talking to him and I was listening in. He said he remembered MSB, but you never know. He also told us is was dark outside and that he likes the dark.

    ReplyDelete
  4. .

    All the better to leap out and scare people in.

    ReplyDelete
  5. Wait. You really think Sunfall Festival is a better band than the Beatles? Becasue if so, well, I just don't know if I can trust your judgement anymore.

    ReplyDelete
  6. .

    Think about it love: how many chicks are in the Beatles?

    ReplyDelete
  7. recessioncone1/16/2007 9:23 PM

    Bang Bang Bang was one of those albums that I hated at first, because it wasn't what I was expecting. But it was very tricksy and kept suggesting itself (very quietly) when I was distracted and needing to choose new music. After a few months I realized I was listening to it quite frequently for an album that I hated.

    ReplyDelete
  8. .

    Let $ C$ be a convex set in $ \mathbb{R}^n$. If $ C$ is bounded, then for any $ x\in C$, any ray emanating from $ x$ will eventually “exit” $ C$ (that is, there is a point $ z$ on the ray such that $ z\notin C$). If $ C$ is unbounded, however, then there exists a point $ x\in C$, and a ray $ \rho$ emanating from $ x$ such that $ \rho\subseteq C$. A direction $ d$ in $ C$ is a point in $ \mathbb{R}^n$ such that for any $ x\in C$, the ray $ \lbrace x+rd\mid r\ge 0\rbrace$ is also in $ C$ (a subset of $ C$).

    The recession cone of $ C$ is the set of all directions in $ C$, and is denoted by denoted by $ 0^{+}C$. In other words,
    $\displaystyle 0^{+}C=\lbrace d\mid x+rd\in C,\ \forall x\in C,\ \forall r\ge 0\rbrace.$

    If a convex set $ C$ is bounded, then the recession cone of $ C$ is pretty useless; it is $ \lbrace 0\rbrace$. The converse is not true, as illustrated by the convex set
    $\displaystyle C=\lbrace (x,y)\mid 0\le x< 1,\ y \ge 1\rbrace \cup \lbrace (x,y)\mid 0\le x\le 1,\ 0\le y\le 1\rbrace.$
    Clearly, $ C$ is not bounded but $ 0^{+}C=\lbrace 0\rbrace$. However, if the additional condition that $ C$ is closed is imposed, then we recover the converse.

    Here are some other examples of recession cones of unbounded convex sets:

    * If $ C=\lbrace (x,y)\mid \vert x\vert \le y\rbrace$, then $ 0^{+}C=C$.
    * If $ C=\lbrace (x,y)\mid \vert x\vert < y\rbrace$, then $ 0^{+}C=\overline{C}$, the closure of $ C$.
    * If $ C=\lbrace (x,y)\mid \vert x\vert^{n}\le y, n>1\rbrace$, then $ 0^{+}C=\lbrace (0,y)\mid y\ge 0\rbrace$.

    ReplyDelete
  9. .

    (whatever that means)

    ReplyDelete
  10. I know their drummer. Or at least I used to. Neener. I met a guy who swore I looked & sounded just like Amy. Only he's never heard me sing. That would've shattered the illusion RIGHT away.

    ReplyDelete
  11. .

    Was that his way of hitting on you?

    ReplyDelete